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"Chapter 1 - The Op Amp's Place in the World" - HTL Wien 10

"Chapter 1 - The Op Amp's Place in the World" - HTL Wien 10

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S<strong>in</strong>e Wave Oscillator Circuits<br />

15-<strong>10</strong><br />

<strong>the</strong> output and Z 1, V TEST is applied to Z 1, and V OUT is calculated. <strong>The</strong> positive feedback<br />

voltage, V +, is calculated first <strong>in</strong> Equations 15–6 through 15–8. Equation 15–6 shows <strong>the</strong><br />

simple voltage divider at <strong>the</strong> non<strong>in</strong>vert<strong>in</strong>g <strong>in</strong>put. Each term is <strong>the</strong>n multiplied by (R 2C 2s<br />

+ 1) and divided by R 2 to get Equation 15–7.<br />

V VTEST Z4 V<br />

Z3 Z4 TEST <br />

R2 R2C2s1 R 1 1<br />

C1s V <br />

V TEST<br />

<br />

<br />

1<br />

1 R1C2S R1 R2 1<br />

R2C1s C2 C1 R 2<br />

R 2 C 2 s1 <br />

<br />

(15–6)<br />

(15–7)<br />

Substitute s = jω 0, where ω 0 is <strong>the</strong> oscillation frequency, ω 1 = 1/R 1C 2, and ω 2 = 1/R 2C 1<br />

to get Equation 15–8.<br />

V <br />

V TEST<br />

<br />

1 R 1<br />

R 2<br />

1<br />

C2 j C1 0 <br />

1 2 0 (15–8)<br />

Some <strong>in</strong>terest<strong>in</strong>g relationships now become apparent. <strong>The</strong> capacitor <strong>in</strong> <strong>the</strong> zero, represented<br />

by ω 1, and <strong>the</strong> capacitor <strong>in</strong> <strong>the</strong> pole, represented by ω 2, must each contribute 90<br />

of phase shift toward <strong>the</strong> 180 required for oscillation at ω 0. This requires that C 1 = C 2 and<br />

R 1 = R 2. Sett<strong>in</strong>g ω 1 and ω 2 equal to ω 0 cancels <strong>the</strong> frequency terms, ideally remov<strong>in</strong>g any<br />

change <strong>in</strong> amplitude with frequency s<strong>in</strong>ce <strong>the</strong> pole and zero negate one ano<strong>the</strong>r. An overall<br />

feedback factor of β = 1/3 is <strong>the</strong> result (Equation 15–9).<br />

V <br />

V TEST<br />

<br />

1<br />

1 R C<br />

R C j 0 0 <br />

1<br />

3 j 0<br />

0 0<br />

0<br />

1<br />

3<br />

(15–9)<br />

<strong>The</strong> ga<strong>in</strong> of <strong>the</strong> negative feedback portion, A, of <strong>the</strong> circuit must <strong>the</strong>n be set such that ⎟Aβ⎟<br />

= 1, requir<strong>in</strong>g A = 3. R F must be set to twice <strong>the</strong> value of R G to satisfy <strong>the</strong> condition. <strong>The</strong><br />

op amp <strong>in</strong> Figure 15–7 is s<strong>in</strong>gle supply, so a dc reference voltage, V REF, must be applied<br />

to bias <strong>the</strong> output for full-scale sw<strong>in</strong>g and m<strong>in</strong>imal distortion. Apply<strong>in</strong>g V REF to <strong>the</strong> positive<br />

<strong>in</strong>put through R 2 restricts dc current flow to <strong>the</strong> negative feedback leg of <strong>the</strong> circuit. V REF<br />

was set at 0.833V to bias <strong>the</strong> output at <strong>the</strong> midrail of <strong>the</strong> s<strong>in</strong>gle supply, rail-to-rail <strong>in</strong>put and<br />

output amplifier, or 2.5 V. See Cahpter 4 for details on dc bias<strong>in</strong>g s<strong>in</strong>gle-supply op amps.<br />

V REF is shorted to ground for split supply applications.<br />

<strong>The</strong> f<strong>in</strong>al circuit is shown <strong>in</strong> Figure 15–8, with component values selected to provide an<br />

oscillation frequency of ω 0 = 2πf 0, where f 0 = 1/(2πRC) = 15.9 kHz. <strong>The</strong> circuit oscillated<br />

at 1.57 kHz due to slightly vary<strong>in</strong>g component values with 2% distortion. This high value<br />

is due to <strong>the</strong> extensive clipp<strong>in</strong>g of <strong>the</strong> output signal at both supply rails, produc<strong>in</strong>g several

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